Question

A monkey standing on the ground wants to climb to the top of the vertical pole of ${\text{13inch}}$ tall. He climbs ${\text{5inch}}$ in $1\sec$ and then slips downwards ${\text{3inch}}$ in the next second. He again climbs ${\text{5inch}}$ in $1\sec$ and slips by ${\text{3inch}}$ in the next second and so on. How much time will the monkey take to reach the top of the pole?
$\left( a \right)\,\,7\sec$
$\left( b \right)\,\,9\sec$
$\left( c \right)\,\,11\sec$
$\left( d \right)\,\,13\sec$

Answer 1

We know the travelling distance of monkey in the first two second, as in the question it is given that the in every one second the monkey go up by ${\text{5inch}}$ and in the next one second the monkey come down by ${\text{3inch}}$ . So from this we can calculate the time will the monkey take to reach the top of the pole.

Complete step by step answer: From the question, we know that the monkey is climbing the pole for ${\text{5inch}}$ in $1\sec$ , but in the successive second the money climbs down ${\text{3inch}}$on the pole.
So we can obtain the effective distance covered in the first two second, which is the one cycle of the monkey including climbing up and down. We have to assume the direction of climbing up and down, so climb up distance is taken as positive while climb down distance is taken as negative.
The effective distance in the first $2\sec$ of the monkey, which will make one cycle:
$\Rightarrow d = 5{\text{inch}} + \left( {{\text{ - 3inch}}} \right)$
And on solving the above solution, we get
$\Rightarrow d = 2{\text{inch}}$
Thus, the monkey climbs up ${\text{2inch}}$ in the first $2\sec$ and in other $2\sec$ , it climbs up ${\text{4inch}}$ and so on.
After $10\sec$ of climbing up and down, the monkey reaches to a height of ${\text{10inch}}$ , but on the ${\text{1}}{{\text{1}}^{{\text{th}}}}{\text{sec}}$ , the monkey reached to the top of the pole by climbing the remaining ${\text{3inch}}$ .
Thus, the time required by the monkey to reach the top of the pole is $11\sec$ .

Therefore, option $\left( c \right)$ is the correct option.

Note: The monkey moves up and down, make sure to take the one distance as negative. According to the question, we only need to find the time at which the monkey reaches to the top of the pole. So we can take the minimum time required.